A 4.75-g bullet is fired with a velocity of 120.0 m/s toward a 20.0-kg stationary solid block resting on a frictionless surface. What is the change in momentum of the bullet if it is embedded in the block?
2006-11-26 09:07:40 · 3 answers · asked by lindi 2 in Science & Mathematics ➔ Physics
You have to apply the law of onservation of momentum.
Inintial momentum = 0.00475 kg x 120 m/s = 0.570 kg m/s
Let v be the common velocity of the block and bullet
Final mommentum = [20+0.00475] x v
v = 0.570/20.00475 = 120 = 0.0285 m/s
Final momentum of bullet = 0.00475 x 0.0285 = 0.000135
Change in momentum 0.570 -0.000135 = 0.569865 kg m/s
2006-11-26 10:48:55 · answer #1 · answered by Let'slearntothink 7 · 1⤊ 0⤋
It would seem to me that after it is embedded in the block it's velocity would be zero so the change in momentum would be the velocity of 120. m/s.
2006-11-26 17:20:30 · answer #2 · answered by Anonymous · 0⤊ 1⤋
RIGHT NOW IM FAILING PHYSICS WISH I COULD HELP THO LOL
2006-11-26 17:15:03 · answer #3 · answered by stephie 3 · 0⤊ 1⤋